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JEE Main & Advanced Physics Two Dimensional Motion Question Bank Self Evaluation Test - Motion in a Plane

  • question_answer
    An object is projected with a velocity of \[20\text{ }m/s\] making an angle of \[45{}^\circ \] with horizontal. The equation for the trajectory is \[h=Ax-B{{x}^{2}}\] where h is height, x is horizontal distance, A and B are constants. The ratio A: B is \[\left( \text{g = 10 m}{{\text{s}}^{-2}} \right)\]

    A) \[1\text{ }:\text{ }5\]     

    B)        \[5\text{ }:\text{ }1~~\]

    C) \[1\text{ }:\text{ }40\]   

    D)        \[40\text{ }:\text{ }1\]

    Correct Answer: D

    Solution :

    [d] Given \[h=Ax-\text{B}{{\text{x}}^{\text{2}}}\], on comparing with \[y=\text{x}\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\], we get \[A=\tan \theta =tan45{}^\circ =1,\] and \[\text{B}\,\,\text{=}\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }=\frac{10}{2\times {{20}^{2}}\times {{\cos }^{2}}45{}^\circ }=\frac{1}{40}\] \[\therefore \,\,\,\,\frac{\text{A}}{\text{B}}=40.\]


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